（light OJ 1005） Rooks dp

http://www.lightoj.com/volume_showproblem.php?problem=1005

 

 

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Time Limit: 1 second(s)

Memory Limit: 32 MB

A rook is a piece used in the game of chess which is played on a board of square grids. A rook can only move vertically or horizontally from its current position and two rooks attack each other if one is on the path of the other. In the following figure, the dark squares represent the reachable locations for rook R₁ from its current position. The figure also shows that the rook R₁ and R₂ are in attacking positions where R₁ and R₃ are not. R₂ and R₃ are also in non-attacking positions.

Now, given two numbers n and k, your job is to determine the number of ways one can put k rooks on an n x n chessboard so that no two of them are in attacking positions.

Input

Input starts with an integer T (≤ 350), denoting the number of test cases.

Each case contains two integers n (1 ≤ n ≤ 30) and k (0 ≤ k ≤ n²).

Output

For each case, print the case number and total number of ways one can put the given number of rooks on a chessboard of the given size so that no two of them are in attacking positions. You may safely assume that this number will be less than 10¹⁷.

Sample Input	Output for Sample Input
8 1 1 2 1 3 1 4 1 4 2 4 3 4 4 4 5	Case 1: 1 Case 2: 4 Case 3: 9 Case 4: 16 Case 5: 72 Case 6: 96 Case 7: 24 Case 8: 0

 

 

 

 

dp[n][k] 可以由三种状态转化而来：

1.  第 n 层不放象棋  dp[n-1][k]

2.  第 n 层放一个象棋， 可以放象棋的位置为图中蓝色的位置，即(2*(i-j)+1)*dp[i-1][j-1]

3.  第 n 层放两个象棋， 可以放象棋的位置为图中浅蓝色的位置 (i-j+1)*(i-j+1)*dp[i-1][j-2]

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

#define N 1100

#define met(a,b) (memset(a,b,sizeof(a)))
typedef long long LL;

LL dp[N][N];

///dp[n][k] 代表前n*n的矩阵中，放k个象棋的方法数

void Init()
{
    int i, j;

    for(i=0; i<=30; i++)
        dp[i][0] = 1;

    for(i=1; i<=30; i++)
    {
        for(j=1; j<=i; j++)
        {
            dp[i][j] += dp[i-1][j];
            dp[i][j] += (2*(i-j)+1)*dp[i-1][j-1];
            if(j>=2)
            dp[i][j] += (i-j+1)*(i-j+1)*dp[i-1][j-2];
       }
    }

}

int main()
{
    int T, iCase=1;
    scanf("%d", &T);

    Init();
    while(T--)
    {
        int n, k;

        scanf("%d%d", &n, &k);

        printf("Case %d: %lld\n", iCase++, dp[n][k]);
    }
    return 0;
}